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(B) The energy stored in the inductor is given by $U = \frac{1}{2} LI^2$.Substituting the given values,$U = \frac{1}{2} 	imes 1 	imes 1^2 = 0.5 	ex

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(B) The energy stored in the inductor is given by $U = \frac{1}{2} LI^2$.Substituting the given values,$U = \frac{1}{2} 	imes 1 	imes 1^2 = 0.5 	ext{ J}$.When the circuit is broken,this energy is transferred to the capacitor to prevent sparking.The energy stored in a capacitor is $U = \frac{1}{2} CV^2$.Equating the two energies: $\frac{1}{2} CV^2 = \frac{1}{2} LI^2$.Solving for $C$: $C = L \left( \frac{I}{V} ight)^2$.Substituting the values: $C = 1 	imes \left( \frac{1}{500} ight)^2 = \frac{1}{250000} 	ext{ F}$.$C = 4 	imes 10^{-6} 	ext{ F} = 4 \mu 	ext{ F}$.

$A$ circuit having a self-inductance of $1 \text{ H}$ carries a current of $1 \text{ A}$. To prevent sparking when the circuit is broken,a capacitor which can withstand $500 \text{ V}$ is connected across the switch. What is the minimum value of the capacitance of the capacitor (in $\mu \text{ F}$)?

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